For any $(\kappa,\lambda)$ it's easy to construct a real closed field with a $(\kappa,\lambda)$-cut. Let $L$ be the linear order with a increasing $\kappa$-chain followed by a decreasing $\lambda$-chain, i.e., $\kappa+\lambda^*$. Let $F$ be the field ${\mathbb Q}(X_l:l\in L)$ where $(X_l:l\in L)$ are algebraically independent. Order $F$ so that each $X_l$ is positive infinite and every power of $X_l^n$ < $X_j$ whenever $l$<$j$ and let $R$ be the unique real closure of $F$ compatible with the ordering.

The cut of things above the $\kappa$-chain but below the descending $\lambda$-chain is unfilled.

For any $(\kappa,\lambda)$ it's easy to construct a real closed field with a $(\kappa,\lambda)$-cut. Let $L$ be the linear order with a increasing $\kappa$-chain followed by a decreasing $\lambda$-chain, i.e., $\kappa+\lambda^*$. Let $F$ be the field ${\mathbb Q}(X_l:l\in L)$ where $(X_l:l\in L)$ are algebraically independent. Order $F$ so that each $X_l$ is positive infinite and every power of $X_l^n < X_j$ whenever $l< j$ and let $R$ be the unique real closure of $F$ compatible with the ordering.

The cut of things above the $\kappa$-chain but below the descending $\lambda$-chain is unfilled.